What amount of heat should be transferred to the gas

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  • #1
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we havent gone over this in class and its due for points and i have no idea how to do it...


A diatomic gas is confined to a closed container of volume 24.0m3 and pressure 3.03×105 Pa. What amount of heat should be transferred to the gas to increase the pressure by a factor of 6?
 

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  • #2
dextercioby
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Is this all the data from the problem...?Don't they give you the mass of the gas,or any other piece of information?

Daniel.
 
  • #3
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nope, thats all that is given get. :bugeye:
 
  • #4
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isochoric or isothermic

with that data I can say it isn't an adiabatic or isobaric, the it could be isochoric or isothermic process. Do you know if you must get that final pressure with volume variation? or the volume is constant?

Some other stuff: if this is an ideal diatomic gas then the gamma coefficient (that is the division of caloric capacities for constant pressure and volume) is 7/5.
 
  • #5
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Surely the process with that data is isochoric, then the Heat Q that must be provided to system is, without a doubt:

Q=int(n*Cv*dt)

This integral must be evaluted for final temperature an initial one. If the gas is ideal, then:

Q=n*Cv*(Tf-Ti)

where n is the moles of gas Cv the molar caloric capacity at constant volume and Tf-Ti is variation of temperature.

For an ideal diatomic gas, the molar Cv is (5*R)/2 where R is the universal gas constant.

The variation of T must be obtained from the ideal gas state equation:

p*V=n*R*T

for the initial and final coordinates (initial pressure: pi, initial volume:Vi, final pressure: pf, final volume:Vf):

pi*Vi=n*R*Ti
pf*Vf=n*R*Tf

rest the first eq to the second one and you obtain:

(pf*Vf)-(pi*Vi)=n*R*(Tf-Ti)
or:
((pf*Vf)-(pi*Vi))/(n*R)=(Tf-Ti)

substitute this valour of variation of T in the Q expression and you will obtain:

Q=n*Cv*(Tf-Ti)

Q=n*((5*R)/2)*((pf*Vf)-(pi*Vi))/(n*R)

or reducing the terms in eq:

Q=5*((pf*Vf)-(pi*Vi))/2

since pf=6*pi and Vf=Vi (this is the data of your problem)

Q=(5*(5*pi*Vi))/2

Q=(25/2)*pi*Vi

the units are subjective, I recommend to use Joules instead of BTU or other energy and heat units. Convert the data you have (pressures und volumes) to the adhoc units, then substitute the values in the last expression. and you get the answer. if you want to get in Joules convert the pressures to Pascals and the volumes to m^3. since the units of pascal are N/(m^2), pressure times volume is N*m, that unit is J (Joule), then you can easily convert it to BTU or whatever unit you want.
 

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